Asked by rei
                An ore car of mass 41000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 9.6 m lower vertically, is a horizontally situated spring with constant 4.7 × 10^
5 N/m.
The acceleration of gravity is 9.8 m/s^
2. Ignore friction.
How much is the spring compressed in stopping the ore car? Answer in units of m.
            
        5 N/m.
The acceleration of gravity is 9.8 m/s^
2. Ignore friction.
How much is the spring compressed in stopping the ore car? Answer in units of m.
Answers
                    Answered by
            drwls
            
    Gravitational potential energy loss = Spring potential energy increase
M g H = (1/2) k X^2
k = 4.7*10^5 N/m
M = 4.1*10^4 kg
H = 9.6 m
You known what g is.
Solve for X
    
M g H = (1/2) k X^2
k = 4.7*10^5 N/m
M = 4.1*10^4 kg
H = 9.6 m
You known what g is.
Solve for X
                    Answered by
            rei
            
    Thank you!
    
                    Answered by
            rei
            
    I got it wrong. I did exactly what you said. Multiplied m times g times h then multiplied by 2. Sq root that number and got 2778.926411 but it was wrong. 
    
                    Answered by
            drwls
            
    X = sqrt(2 M g H/k)
= 4.54 m
You forgot to divide by k, the spring constant.
    
= 4.54 m
You forgot to divide by k, the spring constant.
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